WebFor finding the radius of 2nd excited state in doubly ionised lithium n=3 because 2nd excited state means 3 orbital &z=3 so _13.6 [3^2/3]=1.587A ertagul 17 Points 2 years ago formula of finding the radius of atom = 0.529 * n^2/ZSo, here the ‘n’ is 2 & ‘Z’ is 3Then, Radius=0.529 * 4/3 =0.705 Ans by it self Answer is =0.705 ertagul 17 Points WebThe energy required to remove an electron from n = 2 state of the hydrogen atom is: Medium. View solution > The ionisation energy of hydrogen atom is 1 3. 6 eV. What will be the ionisation energy of H e +? Medium. View solution > View more. CLASSES AND TRENDING CHAPTER. class 5.
Solved 9- Determine the wavelength necessary to ionize H,
WebAtomic spectra In an atomic emission spectrum, each line represents photons of a specific wavelength emitted when an atom undergoes a transition between two states (ninitial →nfinal). Thus, it is important to realize that each line is identified by two quantum numbers, not one, as many students commonly mistake. WebNov 15, 2007 · transitions from excited states to a final state of n=4. Because Line A corresponds to the lowest energy transition, it must be due to the transition from n=5 to n=4. b. The wavelength at which line B occurs is 216 nm. What is the energy of a single photon of light with this wavelength? J m hc J s m s E h 19 9 34 8 1 9.20 10 216 10 (6.626 10 ... nufree soft wax
Angular momentum in second Bohr orbit of H - Toppr Ask
WebThe total energy of an electron in the first excited state of the hydrogen atom is -3.4 eV. Find out its (i) kinetic energy and (ii) potential energy in this state. Medium. View solution > The energy of the electron in the ground state of hydrogen atom is -13.6 eV. Find the kinetic energy and potential energy of electron in this state. Web1) The radius of Bohr’s first orbit in Li 2+ is: (Aditya vardhan - Adichemistry) 1) 0.0587 pm 2) 17.63 pm 3) 176.3 pm 4) 0.529 pm Logic: Use the following equation to calculate the radius in Angstroms, Å. Then convert the value into picometers. Remember 1 Å=10 2 pm Solution: In picometers, the radius = 0.1763 x 10 2 = 17.63 pm WebSolution The correct option is B 30.6 eV Energy required to move an electron from any state to n =∞ is called the binding energy of that state. So, binding energy = 13.6Z2 n2. For, Li2+, Z=3, n=2 (i.e 1st excited state) Therefore, binding energy = 13.632 22 eV binding energy = 13.69 4 eV Hence binding energy = 30.6 eV Suggest Corrections 0 nufree stainless steel aplicator